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  "body": "\n## Proof\n\nWe show that the counts satisfy the Catalan recursion, and conclude by [[strong induction|0KMVFSHFGZ]].\n\n1. For a natural number $m$, let $T_m$ be the set of [[binary trees|02A8EMC83C]] with exactly $m$ nodes.\n2. A tree of the form $\\operatorname{node}(L, R)$ has at least one node, its root. So the only tree with $0$ nodes is the empty tree, and $T_0$ has $1$ element.\n3. Fix a natural number $n$. By the definition, every tree in $T_{n+1}$ is $\\operatorname{node}(L, R)$ for a unique pair $(L, R)$ of binary trees.\n4. Its number of nodes is $1 + |L| + |R|$, where $|L|$ and $|R|$ are the numbers of nodes. So $|L| = i$ and $|R| = n - i$ for some $0 \\le i \\le n$.\n5. Conversely, for $0 \\le i \\le n$, $L \\in T_i$ and $R \\in T_{n-i}$, the tree $\\operatorname{node}(L, R)$ has $n + 1$ nodes.\n6. By steps 3 to 5, the map $\\operatorname{node}(L, R) \\mapsto (L, R)$ is a bijection from $T_{n+1}$ onto the disjoint union over $0 \\le i \\le n$ of the sets $T_i \\times T_{n-i}$.\n7. We prove by strong induction on $m$ that $T_m$ is finite with $|T_m| = C_m$, where $C_m$ are the [[Catalan numbers|092DRBFKMG]], defined by $C_0 = 1$ and $C_{n+1} = \\sum_{i=0}^{n} C_i\\, C_{n-i}$.\n8. For $m = 0$, step 2 gives $|T_0| = 1 = C_0$.\n9. For $m = n + 1$, assume the claim for all $i \\le n$. By step 6, the [[bijection principle|0GV2573FP4]], the [[rule of product|0VPK4X73Q0]] and the [[rule of sum|0EE2V3CHFK]], $T_{n+1}$ is finite and $|T_{n+1}| = \\sum_{i=0}^{n} C_i\\, C_{n-i} = C_{n+1}$.\n10. Hence the number of binary trees with exactly $n$ nodes is $C_n$ for every natural number $n$.\n\n## Idea\n\nA nonempty binary tree is a root with a left and a right subtree, whose node counts add up to one less. So the counts obey the same recursion and start value as the Catalan numbers.\n",
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